LeetCode:Longest Common Prefix
2015-02-24 22:57
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题目不难,主要是被string类中的erase弄晕了。
class Solution { public: string longestCommonPrefix(vector<string> &strs) { string s; int length; int i,j; if(strs.empty()) return s; s=strs[0];//初始化s为第一个元素 for(i=1;i<strs.size();i++){ for(j=0;j<strs[i].length()&&j<s.length();j++)//从第二个元素开始,分别按照字符顺序比较 { if(s[j]!=strs[i][j])//若不相同,则从下标j开始,把s中j以后的字符串删除 { if(j==0) { s.clear(); return s; } else { s.erase(s.begin()+j,s.end());//begin()+j就等于下标j,end()-j就相当于逆序的下标j break; } } } if(j<s.length()) s.erase(s.begin()+j,s.end()); } return s; } };
以上是通过横向比较,其实采用纵向比较,比较的次数会更少:
class Solution {
public:
string longestCommonPrefix(vector<string> &strs) {
string s;
int length;
int i,j;
if(strs.empty())
return s;
s=strs[0];//初始化s为第一个元素
length=strs[0].length();
for(i=1;i<strs.size();i++)
{
if(strs[i].length()<length)//s的长度截成最小字符串长度
length=strs[i].length();
}
s.erase(s.end()-(s.length()-length),s.end());
for(i=0;i<length;i++){
for(j=1;j<strs.size();j++)
{
if(s[i]!=strs[j][i])
{
if(i==0)
{
s.clear();
return s;
}
else
{
s.erase(s.begin()+i,s.end());
length=s.length();//注意越界问题,更新s的实际长度
break;
}
}
}
}
return s;
}
};
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